1. A blank of 50 mm diameter is to be sheared from a sheet of 2.5 mm thickness. The required radial clearance between the die and the punch is 6% of sheet thickness. The punch and die diameter (mm) for this blanking operation respectively are

1. 1. 50.00 and 50.30

   
   
   

   2. 50.00 and 50.15

   
   
   

   3. 49.70 and 50.00

   
   
   

   4. 49.85 and 50.00

   
   
   

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1. View Hint View Answer Discuss in Forum

   Radial clearance =	6	× t = 0.1 ≤ 1.5 mm
   	100

   
   Total clearance = 2 × 0.15 = 0.3
   Die Dia will be exact = 50 mm
   Clearance will be provided on punch = 50–0.30 = 49.70 mm

   Correct Option: C

   Radial clearance =	6	× t = 0.1 ≤ 1.5 mm
   	100

   
   Total clearance = 2 × 0.15 = 0.3
   Die Dia will be exact = 50 mm
   Clearance will be provided on punch = 50–0.30 = 49.70 mm

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2. Match the following products with the suitable manufacturing process
   

   Product	Manufacturing Process
   P. Toothpaste tube	1. Centrifugal casting
   Q. Metallic pipes	2. Blow moulding
   R. Plastic bottles	3. Rolling
   S. Threaded bolts	4. Impact extrusion

1. 1. P-4,Q-3,R-1,S-2

   
   
   

   2. P-2, Q-1, R-3, S-4

   
   
   

   3. P-4, Q-1, R-2, S-3

   
   
   

   4. P-1, Q-3, R-4, S-2

   
   
   

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1. View Hint View Answer Discuss in Forum

   P – 4, Q – 1, R – 2, S – 3

   Correct Option: C

   P – 4, Q – 1, R – 2, S – 3

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3. It is required to punch a hole of diameter 10 mm on a sheet of thickness 3 mm. The shear strength of the work material is 420 MPa. The required punch force is

1. 1. 19.78 kN

   
   
   

   2. 39.58 kN

   
   
   

   3. 98.9 kN

   
   
   

   4. 359.6 kN

   
   
   

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1. View Hint View Answer Discuss in Forum

   Punching force = πDtτ
   = π × 10 × 3 × 420 = 39.58 KN

   Correct Option: B

   Punching force = πDtτ
   = π × 10 × 3 × 420 = 39.58 KN

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4. In a linearly hardening plastic material, the true stress beyond initial yielding

1. 1. increases linearly with the true strain

   
   
   

   2. decreases linearly with the true strain

   
   
   

   3. first increases linearly and then decreases linearly with the true strain

   
   
   

   4. remain constant

   
   
   

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1. View Hint View Answer Discuss in Forum

   

   Correct Option: A

   

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5. Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness. The shear strength of aluminum is 80 MPa. The minimum punching force required in kN is

1. 1. 2.57

   
   
   

   2. 3.29

   
   
   

   3. 5.03

   
   
   

   4. 6.33

   
   
   

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1. View Hint View Answer Discuss in Forum

   Punching force = πDtτ
   = π × 10 × 2 × 80 = 5.023 KN

   Correct Option: C

   Punching force = πDtτ
   = π × 10 × 2 × 80 = 5.023 KN

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